Athird-degree polynomial function f has real zeros -2, 1⁄2, and 3, and its leading coefficient negative. write an equation for f. sketch the graph of f. how many different polynomials functions are possible for f?

f(x) = -x³ + (3/2)x² + (11/3)x + 3.

Step-by-step explanation:

Given, real zeroes of f(x) are -2, 1/2, 3.

⇒ f(x)  = -(x+2)(x-1/2)(x-3). (- sign because given leading coefficient is negative)

⇒f(x) = -x³ + (3/2)x² + (11/3)x + 3.

as the real zeroes are fixed, only one such polynomial is possible.

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